Answer on Question #68512 Math / Trigonometry
sec alpha = 5/4, find value of tan alpha/1 + tan square alpha
Solution:
sec α = 5 4 \sec \alpha = \frac {5}{4} sec α = 4 5 cos α = 1 sec α = 4 5 , sin α = 1 − cos 2 α = 1 − ( 4 5 ) 2 = 3 5 . \cos \alpha = \frac {1}{\sec \alpha} = \frac {4}{5}, \sin \alpha = \sqrt {1 - \cos^ {2} \alpha} = \sqrt {1 - \left(\frac {4}{5}\right) ^ {2}} = \frac {3}{5}. cos α = sec α 1 = 5 4 , sin α = 1 − cos 2 α = 1 − ( 5 4 ) 2 = 5 3 . tan α = sin α cos α = 3 4 \tan \alpha = \frac {\sin \alpha}{\cos \alpha} = \frac {3}{4} tan α = cos α sin α = 4 3
So
tan α 1 + tan 2 α = 3 4 1 + ( 3 4 ) 2 = 12 25 = 0.48 \frac {\tan \alpha}{1 + \tan^ {2} \alpha} = \frac {\frac {3}{4}}{1 + \left(\frac {3}{4}\right) ^ {2}} = \frac {12}{25} = 0.48 1 + tan 2 α tan α = 1 + ( 4 3 ) 2 4 3 = 25 12 = 0.48
Answer: 12 25 = 0.48 \frac{12}{25} = 0.48 25 12 = 0.48
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